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Thread: Convergence in Distribution

  1. #1
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    Convergence in Distribution

    Let Xn be a sequence of p dimensional random vectors. Show that

    Xn converges in distribution to $\displaystyle N_p(\mu,\Sigma)$ iff $\displaystyle a'X_n$ converges in distribution to $\displaystyle N_1(a' \mu, a' \Sigma a).$

    So I start by finding the MGF to a'Xn:

    $\displaystyle E(e^{(a'X_n)t} = E(e^{(a't)X_n}) = e^{a't \mu + 0.5t^2(a' \Sigma a)}$

    Hence, {a'Xn} converges $\displaystyle N(a' \mu, a' \Sigma a).$ in distribution.

    Is that it, cause I'm not to sure about what I'm supposed to prove.
    Last edited by statmajor; Nov 5th 2009 at 05:49 PM.
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  2. #2
    Moo
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    Hello,

    What you did was finding the MGF of a normal distribution. But all we know is that there is a convergence in distribution. Plus, you can't just change the order of t, a, Xn, because they're matrices/vectors !

    ' denotes the transpose

    Let's suppose $\displaystyle X_n \xrightarrow[n\to\infty]{distribution} N_p(\mu,\Sigma)$

    This means that there is convergence of the MGF's :

    $\displaystyle \forall t\in\mathbb{R}^p ~,~ \mathcal{M}_1(t)=E\left[\exp\left(\langle t,X_n\rangle\right)\right]\xrightarrow[n\to\infty]{} \exp\left(\mu't+\tfrac 12 \cdot t'\Sigma t\right) \quad \quad (*)$ ($\displaystyle \langle,\rangle$ denotes the scalar product)


    Now we have the MGF of $\displaystyle a'X_n$ which is $\displaystyle \mathcal{M}_2(t)=E\left[\exp\left(\langle t,a'X_n\rangle\right)\right]$

    By remembering that $\displaystyle \langle x,y \rangle=x'y$, it's very easy to show that $\displaystyle \langle t,a'X_n\rangle=\langle at,X_n\rangle$


    So we have $\displaystyle \mathcal{M}_2(t)=\mathcal{M}_1(at)$. By (*), it follows that :

    $\displaystyle \mathcal{M}_2(t)\xrightarrow[n\to\infty]{}
    \exp\left(\mu'at+\tfrac 12 \cdot t'a'\Sigma at\right) = \exp\left((a'\mu)'t+\tfrac 12 \cdot t' (a'\Sigma a)t \right)

    $

    which is exactly the MGF of $\displaystyle Z\sim \mathcal{N}_p(a'\mu,a'\Sigma a)$


    Now for the other way of the equivalence, just apply this property to $\displaystyle b(a'X_n)$, where b is such that $\displaystyle ba'=I_p$ (the identity matrix)
    Last edited by Moo; Nov 7th 2009 at 12:55 PM.
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    Thanks a lot for that explanation, which cleared up a lot of things.

    Just have a question, the MGF of $\displaystyle M_2(t)$ is $\displaystyle N_p(a' \mu, a' \Sigma a)$?
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    Quote Originally Posted by statmajor View Post
    Thanks a lot for that explanation, which cleared up a lot of things.

    Just have a question, the MGF of $\displaystyle M_2(t)$ is $\displaystyle N_p(a' \mu, a' \Sigma a)$?
    No, it tends to the MGF of $\displaystyle N_p(a'\mu,a'\Sigma a)$

    Sorry I'll modify a bit my post above so that there is no confusion ^^'


    you have to remember that the MGF's of Xn or a'Xn have their limit equal to the MGF's of $\displaystyle N_p(\mu,\Sigma)$ or $\displaystyle N_p(a'\mu,a'\Sigma a)$
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    Quote Originally Posted by Moo View Post
    you have to remember that the MGF's of Xn or a'Xn have their limit equal to the MGF's of $\displaystyle N_p(\mu,\Sigma)$ or $\displaystyle N_p(a'\mu,a'\Sigma a)$
    That would be a limiting distribution (as my textbook/prof refer to it)?
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    Moo
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    Yes, the limiting distribution of $\displaystyle X_n$ is $\displaystyle N_p(\mu,\Sigma)$ iff the limiting distribution of $\displaystyle a'X_n$ is $\displaystyle N_p(a'\mu,a'\Sigma a)$, as proved above.

    And if D is the limiting distribution of a rv Xn, then the mgf associated to D is the limit of the mgf of Xn.
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  7. #7
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    Again, thanks for all of your help.
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